Connecting Kp and Kc

Introduction

Kp and Kc describe the same gas equilibrium using different composition measures. The ideal-gas law connects each partial pressure to its molar concentration. Their commonly taught conversion depends on the net change in gaseous mole coefficients, so solids and liquids must be excluded from that count.

Core explanation

For an ideal gas species i, p iV = n iRT, so p i = (n i/V)RT = c iRT. Substitute this into every gas factor of a Kp expression. The RT factors collect to exponent Δn gas = sum of gas coefficients on the product side minus sum on the reactant side. In a common textbook numerical convention, Kp = Kc(RT)^Δn gas when R and pressure units are used consistently.

For N₂ + 3H₂ ⇌ 2NH₃, Δn gas = 2 − (1 + 3) = −2. The textbook relation becomes Kp = Kc(RT)⁻². For H₂ + I₂ ⇌ 2HI, Δn gas = 2 − 2 = 0, giving Kp = Kc numerically in the matching simple convention. No calculation of R or T is needed when Δn is zero.

Only gaseous species count. In CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δn gas = 1 − 0 = 1 because the pure solids do not contribute gas moles. Kp and Kc for the gas factor are related by one RT factor in the simple model. Counting solid coefficients would give the wrong exponent and a physically inconsistent relation.

The rigorous thermodynamic constants use dimensionless activities: ideal gas p i/p° and dilute-solution c i/c°. Their conversion includes the standard-state ratio RT c°/p° raised to Δn gas. Thus the bare RT formula presumes a particular numerical convention for Kp and Kc. Introductory exercises usually provide the intended formula and R units; advanced work should keep standard states explicit.

Temperature enters both through RT in the conversion and through the actual temperature dependence of the equilibrium constants. The algebraic relation at one T does not say either constant remains unchanged as T varies. If R is taken as 0.08206 L atm mol⁻¹ K⁻¹, pressure numbers must use atm and concentration numbers mol L⁻¹ for the usual textbook relation.

Step-by-step reasoning

1. Mark only gaseous species in the balanced reaction. 2. Calculate Δn gas from product and reactant gas coefficients. 3. Apply p i = c iRT and collect the RT exponent. 4. Check pressure, concentration and standard-state conventions.

Visual explanation

Draw Kp and Kc fractions side by side. Replace each p term with cRT; circle the leftover RT factors after cancellation.

Real-world analogy

Converting several prices between currencies multiplies each by the exchange factor. If the same number appears above and below a ratio, those factors cancel; Δn counts the unmatched factors.

Real-world example

Ammonia-synthesis equilibrium data may be reported as gas concentrations in one experiment and partial pressures in another. The ideal-gas relation connects their constants at the same temperature and convention.

Why?

Why does Δn = 0 cancel RT? Equal total gaseous coefficient powers appear above and below the expression, so all RT conversion factors cancel algebraically from their ratio.

Common misconception

“Δn includes every coefficient in the equation.” Only gas species enter the Kp–Kc ideal-gas conversion; pure solids and liquids have no gas partial pressure.

Worked example

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Δn gas = 2 − 3 = −1. If an exercise gives Kc = 4.00 and RT = 25.0 in matching pressure/concentration numerical units, Kp = 4.00(25.0)⁻¹ = 0.160 in that convention. Check the inverse dependence: fewer gaseous product moles yield a negative exponent.

Quick check

1. What is Δn gas for H₂ + I₂ ⇌ 2HI, all gases? Answer: 2 − (1 + 1) = 0, so matching numerical Kp and Kc are equal.

Exam focus

Count gas coefficients only and keep units consistent. If rigorous standard states are used, include their normalization instead of writing a bare RT factor uncritically.

Advanced insight

Real-gas behavior replaces p i with fugacity-based activities, so a simple p i = c iRT substitution may not suffice. The thermodynamic constant remains activity-based while numerical Kp/Kc approximations acquire model-dependent corrections.

Summary

The ideal-gas identity p i = c iRT produces a Kp/Kc conversion governed by Δn gas. The common Kp = Kc(RT)^Δn formula requires matching numerical conventions and only gaseous stoichiometry.

Practice questions

1. Find Δn gas for N₂ + 3H₂ ⇌ 2NH₃. Answer: −2 because two gas moles appear on the product side and four on the reactant side. 2. Do CaCO₃(s) and CaO(s) contribute to Δn gas in their decomposition? Answer: No. They are solids, not gaseous species. 3. What happens to the RT conversion factor if Δn gas = 0? Answer: It becomes one, so the matching textbook numerical constants coincide.